An air source heat pump has a cold side of -25 C and a hot side of 50 C. What is the COP of heating?

Question

BlackTom AI · Accepted Answer

First, recall the formula for the heating COP of a heat pump in terms of temperatures: COP_heating ≈ T_hot / (T_hot − T_cold), where temperatures are in Kelvin.

Option a. 4,3
This value is consistent with the calculation using T_hot = 50°C → 323 K and T_cold = −25°C → 248 K. COP_heating ≈ 323 / (323 − 248) = 323 / 75 ≈ 4.31, which aligns with “4,3” when using European decimal notation.

Option b. 11,3
A COP of about 11.3 is far higher than what the thermodynamic limit for the given temperature difference would allow under the Carnot-like idealization. This would imply an extremely efficient device, which is not feasible given a 75 K temperature lift.

Option c. 14,7
Similarly, 14.7 would require an even larger COP than the Carnot limit permits for a 75 K lift. This suggests a misapplication of the formula or a miscalculation of temperatures.

Option d. 3,3
A COP of 3.3 would correspond to a larger temperature lift or lower hot-side temperature in the ideal formula, but with the actual numbers (323 K hot side, 248 K cold side), the calculated COP is notably higher than 3.3, indicating this value is too low for the given conditions.

In summary, the calculation using T_hot = 323 K and T_cold = 248 K yields a COP near 4.31, which corresponds to approximately 4,3 in the given notation. The other options are inconsistent with the thermodynamic relationship and the specific temperature difference provided.

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An air source heat pump has a cold side of -25 C and a hot side of 50 C. What is the COP of heating?

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